Solution Manual for Mechanics of Materials SI, 9th Edition

© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–1. An air-filled rubber ball has a diameter of 6 in. If the air pressure within it is increased until the ball’s diameter becomes 7 in., determine the average normal strain in the rubber. d0 = 6 in. d = 7 in. P = pd – pd0 7 – 6 = = 0.167 in.>in. pd0 6 Ans. Ans: P = 0.167 in.>in. 105 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip. L0 = 15 in. L = p(5 in.) P = L – L0 5p – 15 = = 0.0472 in.>in. L0 15 Ans. Ans: P = 0.0472 in.>in. 106 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD. D E 4m P A B 3m C 2m 2m ¢LBD ¢LCE = 3 7 3 (10) = 4.286 mm 7 ¢LCE 10 = = 0.00250 mm>mm PCE = L 4000 ¢LBD = PBD = Ans. ¢LBD 4.286 = = 0.00107 mm>mm L 4000 Ans. Ans: PCE = 0.00250 mm>mm, PBD = 0.00107 mm>mm 107 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–4. The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin B through an angle of 2°. Determine the average normal strain developed in each wire. The wires are unstretched when the lever is in the horizontal position. G 200 mm H dA = 200(0.03491) = 6.9813 mm dC = 300(0.03491) = 10.4720 mm dD = 500(0.03491) = 17.4533 mm Average Normal Strain: The unstretched length of wires AH, CG, and DF are LAH = 200 mm, LCG = 300 mm, and LDF = 300 mm. We obtain dA 6.9813 = = 0.0349 mm>mm LAH 200 dC 10.4720 = 0.0349 mm>mm = (Pavg)CG = LCG 300 (Pavg)AH = (Pavg)DF = Ans. Ans. dD 17.4533 = = 0.0582 mm>mm LDF 300 Ans. 108 E C 200 mm 2° bp rad = 0.03491 rad. 180 Since u is small, the displacements of points A, C, and D can be approximated by 200 mm 300 mm 300 mm B A Geometry: The lever arm rotates through an angle of u = a F D © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–5. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire. C 300 mm 30⬚ 30⬚ 300 A P mm B œ LAC = 23002 + 22 – 2(300)(2) cos 150° = 301.734 mm PAC = PAB = œ – LAC LAC 301.734 – 300 = = 0.00578 mm>mm LAC 300 Ans. Ans: PAC = PAB = 0.00578 mm>mm 109 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–6. The rubber band of unstretched length 2r0 is forced down the frustum of the cone. Determine the average normal strain in the band as a function of z. r0 z h 2r0 Geometry: Using similar triangles shown in Fig. a, h¿ h¿ + h = ; r0 2r0 h¿ = h Subsequently, using the result of h¿ r0 r = ; z+h h r = r0 (z + h) h Average Normal Strain: The length of the rubber band as a function of z is 2pr0 (z +h). With L0 = 2r0, we have L = 2pr = h L – L0 Pavg = = L0 2pr0 (z + h) – 2r0 h p = (z + h) – 1 2r0 h Ans. Ans: Pavg = 110 p (z + h) – 1 h © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–7. The pin-connected rigid rods AB and BC are inclined at u = 30° when they are unloaded. When the force P is applied u becomes 30.2°. Determine the average normal strain developed in wire AC. P B u u 600 mm A C Geometry: Referring to Fig. a, the unstretched and stretched lengths of wire AD are LAC = 2(600 sin 30°) = 600 mm LAC ¿ = 2(600 sin 30.2°) = 603.6239 mm Average Normal Strain: (Pavg)AC = LAC ¿ – LAC 603.6239 – 600 = = 6.04(10 – 3) mm>mm LAC 600 Ans. Ans: (Pavg)AC = 6.04(10 – 3) mm>mm 111 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. u *2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched. D P 300 mm B 300 mm A C 400 mm AB = 24002 + 3002 = 500 mm AB¿ = 24002 + 3002 – 2(400)(300) cos 90.3° = 501.255 mm PAB = AB¿ – AB 501.255 – 500 = AB 500 = 0.00251 mm>mm Ans. 112 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–9. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm> mm, determine the displacement of point D. Originally the cable is unstretched. u D P 300 mm B 300 mm A C 400 mm AB = 23002 + 4002 = 500 mm AB¿ = AB + eABAB = 500 + 0.0035(500) = 501.75 mm 501.752 = 3002 + 4002 – 2(300)(400) cos a a = 90.4185° u = 90.4185° – 90° = 0.4185° = ¢ D = 600(u) = 600( p (0.4185) rad 180° p )(0.4185) = 4.38 mm 180° Ans. Ans: ¢ D = 4.38 mm 113 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–10. The corners of the square plate are given the displacements indicated. Determine the shear strain along the edges of the plate at A and B. y 0.2 in. A 10 in. D B x 0.3 in. 0.3 in. 10 in. 10 in. C 10 in. 0.2 in. At A: 9.7 u¿ = tan – 1 a b = 43.561° 2 10.2 u¿ = 1.52056 rad (gA)nt = p – 1.52056 2 = 0.0502 rad Ans. At B: f¿ 10.2 = tan – 1 a b = 46.439° 2 9.7 f¿ = 1.62104 rad (gB)nt = p – 1.62104 2 = -0.0502 rad Ans. Ans: (gA)nt = 0.0502 rad, (gB)nt = -0.0502 rad 114 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–11. The corners of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonals AC and DB. y 0.2 in. A 10 in. D B x 0.3 in. 0.3 in. 10 in. 10 in. C 10 in. 0.2 in. For AB: A¿B¿ = 2(10.2)2 + (9.7)2 = 14.0759 in. AB = 2(10)2 + (10)2 = 14.14214 in. PAB = 14.0759 – 14.14214 = -0.00469 in.>in. 14.14214 Ans. 20.4 – 20 = 0.0200 in.>in. 20 Ans. 19.4 – 20 = -0.0300 in.>in. 20 Ans. For AC: PAC = For DB: PDB = Ans: PAB = -0.00469 in.>in., PAC = 0.0200 in.>in., PDB = -0.0300 in.>in. 115 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–12. The piece of rubber is originally rectangular. Determine the average shear strain gxy at A if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines. y 3 mm C D 400 mm A u1 = tan u1 = 2 = 0.006667 rad 300 u2 = tan u2 = 3 = 0.0075 rad 400 gxy = u1 + u2 = 0.006667 + 0.0075 = 0.0142 rad Ans. 116 300 mm B 2 mm x © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–13. The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD. y 3 mm C D 400 mm A 300 mm B 2 mm x AD¿ = 2(400)2 + (3)2 = 400.01125 mm f = tan – 1 a 3 b = 0.42971° 400 AB¿ = 2(300)2 + (2)2 = 300.00667 w = tan – 1 a 2 b = 0.381966° 300 a = 90° – 0.42971° – 0.381966° = 89.18832° D¿B¿ = 2(400.01125)2 + (300.00667)2 – 2(400.01125)(300.00667) cos (89.18832°) D¿B¿ = 496.6014 mm DB = 2(300)2 + (400)2 = 500 mm 496.6014 – 500 = -0.00680 mm>mm 500 400.01125 – 400 PAD = = 0.0281(10 – 3) mm>mm 400 Ans. PDB = Ans. Ans: PDB = -0.00680 mm>mm, PAD = 0.0281(10 – 3) mm>mm 117 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–14. The force P applied at joint D of the square frame causes the frame to sway and form the dashed rhombus. Determine the average normal strain developed in wire AC. Assume the three rods are rigid. D P 200 mm 200 mm E C 3 400 mm A B Geometry: Referring to Fig. a, the stretched length of LAC ¿ of wire AC¿ can be determined using the cosine law. LAC ¿ = 24002 + 4002 – 2(400)(400) cos 93° = 580.30 mm The unstretched length of wire AC is LAC = 24002 + 4002 = 565.69 mm Average Normal Strain: (Pavg)AC = LAC ¿ – LAC 580.30 – 565.69 = = 0.0258 mm>mm LAC 565.69 Ans. Ans: (Pavg)AC = 0.0258 mm>mm 118 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–15. The force P applied at joint D of the square frame causes the frame to sway and form the dashed rhombus. Determine the average normal strain developed in wire AE. Assume the three rods are rigid. D P 200 mm 200 mm E C 3 400 mm A B Geometry: Referring to Fig. a, the stretched length of LAE¿ of wire AE can be determined using the cosine law. LAE¿ = 24002 + 2002 – 2(400)(200) cos 93° = 456.48 mm The unstretched length of wire AE is LAE = 24002 +2002 = 447.21 mm Average Normal Strain: (Pavg)AE = LAE ¿ – LAE 456.48 – 447.21 = 0.0207 mm>mm = LAE 447.21 Ans. Ans: (Pavg)AE = 0.0207 mm>mm 119 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–16. The triangular plate ABC is deformed into the shape shown by the dashed lines. If at A, eAB = 0.0075, PAC = 0.01 and gxy = 0.005 rad, determine the average normal strain along edge BC. y C 300 mm gxy A Average Normal Strain: The stretched length of sides AB and AC are LAC¿ = (1 + ey)LAC = (1 + 0.01)(300) = 303 mm LAB¿ = (1 + ex)LAB = (1 + 0.0075)(400) = 403 mm Also, u = p 180° – 0.005 = 1.5658 rada b = 89.7135° 2 p rad The unstretched length of edge BC is LBC = 23002 + 4002 = 500 mm and the stretched length of this edge is LB¿C¿ = 23032 + 4032 – 2(303)(403) cos 89.7135° = 502.9880 mm We obtain, PBC = LB¿C¿ – LBC 502.9880 – 500 = = 5.98(10 – 3) mm>mm LBC 500 120 Ans. 400 mm B x © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–17. The plate is deformed uniformly into the shape shown by the dashed lines. If at A, gxy = 0.0075 rad., while PAB = PAF = 0, determine the average shear strain at point G with respect to the x¿ and y¿ axes. y¿ y F E x¿ 600 mm gxy C D 300 mm c = 90° – 0.4297° = 89.5703° G A 180° Geometry: Here, gxy = 0.0075 rad a b = 0.4297°. Thus, p rad 300 mm 600 mm B b = 90° + 0.4297° = 90.4297° Subsequently, applying the cosine law to triangles AGF¿ and GBC¿, Fig. a, LGF¿ = 26002 + 3002 – 2(600)(300) cos 89.5703° = 668.8049 mm LGC¿ = 26002 + 3002 – 2(600)(300) cos 90.4297° = 672.8298 mm Then, applying the sine law to the same triangles, sin f sin 89.5703° = ; 600 668.8049 f = 63.7791° sin 90.4297° sin a = ; 300 672.8298 a = 26.4787° Thus, u = 180° – f – a = 180° – 63.7791° – 26.4787° = 89.7422°a p rad b = 1.5663 rad 180° Shear Strain: (gG)x¿y¿ = p p – u = – 1.5663 = 4.50(10 – 3) rad 2 2 Ans. Ans: (gG)x¿y¿ = 4.50(10 – 3) rad 121 x © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–18. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines. y 5 mm 2 mm 2 mm B 4 mm C 300 mm 2 mm D A x 400 mm 3 mm Geometry: For small angles, a = c = 2 = 0.00662252 rad 302 b = u = 2 = 0.00496278 rad 403 Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 A 10 – 3 B rad Ans. (gA)xy = u + c = 0.0116 rad = 11.6 A 10 – 3 B rad Ans. Ans: (gB)xy = 11.6(10 – 3) rad, (gA)xy = 11.6(10 – 3) rad 122 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–19. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines. y 5 mm 2 mm 2 mm B 4 mm C 300 mm 2 mm D A x 400 mm 3 mm Geometry: For small angles, 2 = 0.00496278 rad 403 2 = 0.00662252 rad b = u = 302 Shear Strain: a = c = (gC)xy = a + b = 0.0116 rad = 11.6 A 10 – 3 B rad Ans. (gD)xy = u + c = 0.0116 rad = 11.6 A 10 – 3 B rad Ans. Ans: (gC)xy = 11.6(10 – 3) rad, (gD)xy = 11.6(10 – 3) rad 123 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–20. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB. y 5 mm 2 mm 2 mm B 4 mm C 300 mm 2 mm D A 400 mm 3 mm Geometry: AC = DB = 24002 + 3002 = 500 mm DB¿ = 24052 + 3042 = 506.4 mm A¿C¿ = 24012 + 3002 = 500.8 mm Average Normal Strain: PAC = A¿C¿ – AC 500.8 – 500 = AC 500 = 0.00160 mm>mm = 1.60 A 10 – 3 B mm>mm PDB = Ans. DB¿ – DB 506.4 – 500 = DB 500 = 0.0128 mm>mm = 12.8 A 10 – 3 B mm>mm Ans. 124 x © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–21. The rectangular plate is deformed into the shape of a parallelogram shown by the dashed lines. Determine the average shear strain gxy at corners A and B. y 5 mm D C 300 mm 5 mm A B x 400 mm Geometry: Referring to Fig. a and using small angle analysis, u = 5 = 0.01667 rad 300 f = 5 = 0.0125 rad 400 Shear Strain: Referring to Fig. a, (gA)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad Ans. (gB)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad Ans. Ans: (gA)xy = 0.0292 rad, (gB)xy = 0.0292 rad 125 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–22. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the shear strain, gxy, at A. y 45⬚ 800 mm 45⬚ x¿ A 45⬚ A¿ 5 mm 800 mm x L = 28002 + 52 – 2(800)(5) cos 135° = 803.54 mm sin 135° sin u = ; 803.54 800 gxy = u = 44.75° = 0.7810 rad p p – 2u = – 2(0.7810) 2 2 = 0.00880 rad Ans. Ans: gxy = 0.00880 rad 126 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–23. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px along the x axis. y 45⬚ 800 mm 45⬚ x¿ A 45⬚ A¿ 5 mm 800 mm x L = 28002 + 52 – 2(800)(5) cos 135° = 803.54 mm Px = 803.54 – 800 = 0.00443 mm>mm 800 Ans. Ans: Px = 0.00443 mm>mm 127 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–24. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px¿ along the x¿ axis. y 45⬚ 800 mm 45⬚ x¿ A 45⬚ 800 mm x L = 800 cos 45° = 565.69 mm Px¿ = 5 = 0.00884 mm>mm 565.69 Ans. 128 A¿ 5 mm © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y 2–25. The square rubber block is subjected to a shear strain of gxy = 40(10 – 6)x + 20(10 – 6)y, where x and y are in mm. This deformation is in the shape shown by the dashed lines, where all the lines parallel to the y axis remain vertical after the deformation. Determine the normal strain along edge BC. Shear Strain: Along edge DC, y = 400 mm. Thus, (gxy)DC = 40(10 – 6)x + 0.008. dy Here, = tan (gxy)DC = tan 340(10 – 6)x + 0.0084. Then, dx dc C A B 400 mm 300 mm dy = L0 D dc = – tan [40(10 – 6)x + 0.008]dx L0 300 mm e ln cos c 40(10 – 6)x + 0.008 d f ` 40(10 – 6) 0 1 x 300 mm = 4.2003 mm Along edge AB, y = 0. Thus, (gxy)AB = 40(10 – 6)x. Here, dy = tan (gxy)AB = dx tan [40(10 – 6)x]. Then, 300 mm dB dy = L0 dB = – L0 tan [40(10 – 6)x]dx 300 mm e ln cos c 40(10 – 6)x d f ` 40(10 – 6) 0 1 = 1.8000 mm Average Normal Strain: The stretched length of edge BC is LB¿C¿ = 400 + 4.2003 – 1.8000 = 402.4003 mm We obtain, (Pavg)BC = LB¿C¿ – LBC 402.4003 – 400 = = 6.00(10 – 3) mm>mm LBC 400 Ans. Ans: (Pavg)BC = 6.00(10 – 3) mm>mm 129 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–26. The square plate is deformed into the shape shown by the dashed lines. If DC has a normal strain Px = 0.004, DA has a normal strain Py = 0.005 and at D, gxy = 0.02 rad, determine the average normal strain along diagonal CA. y y¿ x¿ 600 mm A¿ B¿ B A E 600 mm D C C¿ x Average Normal Strain: The stretched length of sides DA and DC are LDC¿ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm LDA¿ = (1 + Py)LDA = (1 + 0.005)(600) = 603 mm Also, a = p 180° – 0.02 = 1.5508 rada b = 88.854° 2 p rad Thus, the length of C¿A¿ can be determined using the cosine law with reference to Fig. a. LC¿A¿ = 2602.42 + 6032 – 2(602.4)(603) cos 88.854° = 843.7807 mm The original length of diagonal CA can be determined using Pythagorean’s theorem. LCA = 26002 + 6002 = 848.5281 mm Thus, (Pavg)CA = LC¿A¿ – LCA 843.7807 – 848.5281 = -5.59(10 – 3) mm>mm Ans. = LCA 848.5281 Ans: (Pavg)CA = -5.59(10 – 3) mm>mm 130 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–27. The square plate ABCD is deformed into the shape shown by the dashed lines. If DC has a normal strain Px = 0.004, DA has a normal strain Py = 0.005 and at D, gxy = 0.02 rad, determine the shear strain at point E with respect to the x¿ and y¿ axes. y y¿ x¿ 600 mm A¿ B¿ B A 600 mm D E C C¿ x Average Normal Strain: The stretched length of sides DC and BC are LDC¿ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm LB¿C¿ = (1 + Py)LBC = (1 + 0.005)(600) = 603 mm Also, a = 180° p – 0.02 = 1.5508 rada b = 88.854° 2 p rad f = 180° p + 0.02 = 1.5908 rada b = 91.146° 2 p rad Thus, the length of C¿A¿ and DB¿ can be determined using the cosine law with reference to Fig. a. LC¿A¿ = 2602.42 + 6032 – 2(602.4)(603) cos 88.854° = 843.7807 mm LDB¿ = 2602.42 + 6032 – 2(602.4)(603) cos 91.146° = 860.8273 mm Thus, LE¿A¿ = LC¿A¿ = 421.8903 mm 2 LE¿B¿ = LDB¿ = 430.4137 mm 2 Using this result and applying the cosine law to the triangle A¿E¿B¿ , Fig. a, 602.42 = 421.89032 + 430.41372 – 2(421.8903)(430.4137) cos u u = 89.9429°a p rad b = 1.5698 rad 180° Shear Strain: (gE)x¿y¿ = p p – u = – 1.5698 = 0.996(10 – 3) rad 2 2 Ans. Ans: (gE)x¿y¿ = 0.996(10 – 3) rad 131 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–28. The wire is subjected to a normal strain that is 2 P ⫽ (x/L)e–(x/L) 2 defined by P = (x>L)e – (x>L) . If the wire has an initial x length L, determine the increase in its length. x L L ¢L = 1 – (x>L)2 xe dx L L0 = -L c = 2 L e – (x>L) L d = 31 – (1>e)4 2 2 0 L 3e – 14 2e Ans. 132 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–29. The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal AC, and the average shear strain at corner A. y 6 mm 400 mm 2 mm 2 mm 6 mm C D 300 mm 2 mm A Geometry: The unstretched length of diagonal AC is x B 400 mm 3 mm LAC = 2300 + 400 = 500 mm 2 2 Referring to Fig. a, the stretched length of diagonal AC is LAC¿ = 2(400 + 6)2 + (300 + 6)2 = 508.4014 mm Referring to Fig. a and using small angle analysis, f = 2 = 0.006623 rad 300 + 2 a = 2 = 0.004963 rad 400 + 3 Average Normal Strain: Applying Eq. 2, (Pavg)AC = LAC¿ – LAC 508.4014 – 500 = = 0.0168 mm>mm LAC 500 Ans. Shear Strain: Referring to Fig. a, (gA)xy = f + a = 0.006623 + 0.004963 = 0.0116 rad Ans. Ans: (Pavg)AC = 0.0168 mm>mm, (gA)xy = 0.0116 rad 133 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–30. The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal BD, and the average shear strain at corner B. y 6 mm 400 mm 2 mm 2 mm 6 mm C D 300 mm 2 mm A Geometry: The unstretched length of diagonal BD is x B 400 mm 3 mm LBD = 23002 + 4002 = 500 mm Referring to Fig. a, the stretched length of diagonal BD is LB¿D¿ = 2(300 + 2 – 2)2 + (400 + 3 – 2)2 = 500.8004 mm Referring to Fig. a and using small angle analysis, 2 = 0.004963 rad 403 3 = 0.009868 rad a = 300 + 6 – 2 f = Average Normal Strain: Applying Eq. 2, (Pavg)BD = LB¿D¿ – LBD 500.8004 – 500 = = 1.60(10 – 3) mm>mm Ans. LBD 500 Shear Strain: Referring to Fig. a, (gB)xy = f + a = 0.004963 + 0.009868 = 0.0148 rad Ans. Ans: (Pavg)BD = 1.60(10 – 3) mm>mm, (gB)xy = 0.0148 rad 134 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–31. The nonuniform loading causes a normal strain in the shaft that can be expressed as Px = kx2, where k is a constant. Determine the displacement of the end B. Also, what is the average normal strain in the rod? L A B x d(¢x) = Px = kx2 dx L (¢x)B = 3 2 L0 kx = kL 3 Ans. kL3 (¢x)B kL2 3 (Px)avg = = = L L 3 Ans. Ans: 3 (¢x)B = 135 kL kL2 , (Px)avg = 3 3 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–32 The rubber block is fixed along edge AB, and edge CD is moved so that the vertical displacement of any point in the block is given by v(x) = (v0>b3)x3. Determine the shear strain gxy at points (b>2, a>2) and (b, a). y v (x) A v0 D a B Shear Strain: From Fig. a, b dv = tan gxy dx 3v0 2 x = tan gxy b3 gxy = tan – 1 a 3v0 2 x b b3 Thus, at point (b> 2, a> 2), gxy = tan – 1 c 3v0 b 2 a b d b3 2 3 v0 = tan – 1 c a b d 4 b Ans. and at point (b, a), gxy = tan – 1 c b3 (b2) d = tan – 1 c 3 a v0 bd b 3v0 C Ans. 136 x © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–33. The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements uA and vB, respectively, determine the normal strain in the fiber when it is in position A¿B¿ . y B¿ vB B L A uA A¿ u x Geometry: LA¿B¿ = 2(L cos u – uA)2 + (L sin u + vB)2 = 2L2 + u2A + v2B + 2L(vB sin u – uA cos u) Average Normal Strain: LA¿B¿ – L PAB = L = A 1 + 2(vB sin u – uA cos u) u2A + v2B + – 1 L L2 Neglecting higher terms u2A and v2B 1 PAB = B 1 + 2(vB sin u – uA cos u) 2 R – 1 L Using the binomial theorem: PAB = 1 + = 2uA cos u 1 2vB sin u ¢ ≤ + … – 1 2 L L vB sin u uA cos u L L Ans. Ans. PAB = 137 vB sin u uA cos u L L © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–34. If the normal strain is defined in reference to the final length, that is, P¿n = lim ¿ a p:p ¢s¿ – ¢s b ¢s¿ instead of in reference to the original length, Eq. 2–2 , show that the difference in these strains is represented as a second-order term, namely, Pn – Pn¿ = Pn Pn¿ . PB = ¢S¿ – ¢S ¢S œ = PB – PA ¢S¿ – ¢S ¢S¿ – ¢S ¢S ¢S¿ ¢S¿ 2 – ¢S¢S¿ – ¢S¿¢S + ¢S2 ¢S¢S¿ ¢S¿ 2 + ¢S2 – 2¢S¿¢S = ¢S¢S¿ = = (¢S¿ – ¢S)2 ¢S¿ – ¢S ¢S¿ – ¢S = ¢ ≤¢ ≤ ¢S¢S¿ ¢S ¢S¿ = PA PBœ (Q.E.D) 138